Question

Let G be the set of positive real numbers except 1. Define α∗β = αlnβ. Then:

(a) show that (G,∗) is a group.

(b) Is G abelian? if not, find its center.

(c) Give an automorphism of G.

Answer 1

Let $G$ be the set of positive real numbers except 1. Define $α∗β = α^{\lnβ}$.

(a) Let us show that $(G,∗)$ is a group. Since $α∗β = α^{\lnβ}>0$ and $α^{\lnβ}=1$ if and only if $\ln\beta=0$if and only if $\beta=1,$ we conclude that the operation is defined on the set $G.$

Since $(α∗β)*\gamma = α^{\lnβ}*\gamma=(α^{\lnβ})^{\ln\gamma}=α^{\ln\gamma\lnβ}=α^{\lnβ^{\ln\gamma}} =α^{\ln(β*\gamma)}=α*(β*\gamma),$ we conclude that the operation $*$ is associative.

Since $α∗e = α^{\ln e}=α= e^{\ln α}=e*α$ for any $\alpha\in G,$ we conclude that $e$ is the identity of $G.$ Taking into account that $e^{\frac{1}{\ln \alpha}}*\alpha=(e^{\frac{1}{\ln \alpha}})^{\ln\alpha}=e^{\frac{\ln\alpha}{\ln \alpha}}=e,$ we conclude that $\alpha^{-1}=e^{\frac{1}{\ln \alpha}}\in G,$ and hence $(G,∗)$ is a group.

(b) Since $α∗β = α^{\lnβ}=β^{\ln α}=β ∗α$ for any $α,\beta\in G,$ we conclude that $G$ is abelian group.

(c) For the map $f: G\to G, f(x)=x,$ we have that $f(x*y)=f(x^{\ln y})=x^{\ln y}=x*y=f(x)*f(y),$ and hence $f$ is an automorphism of $G$.